You have 4 cards each containing a number from 1 to 9. You need to judge whether they could operated through *, /, +, -, (, ) to get the value of 24.

Example 1:

Input: [4, 1, 8, 7]
Output: True
Explanation: (8-4) * (7-1) = 24

Example 2:

Input: [1, 2, 1, 2]
Output: False

Note:

1. The division operator / represents real division, not integer division. For example, 4 / (1 - 2/3) = 12.
2. Every operation done is between two numbers. In particular, we cannot use - as a unary operator. For example, with [1, 1, 1, 1] as input, the expression -1 - 1 - 1 - 1 is not allowed.
3. You cannot concatenate numbers together. For example, if the input is [1, 2, 1, 2], we cannot write this as 12 + 12.

思路

有兩種解法，其中一種不是很理解。先來說說另一種方法，Backtracking所有可能。對於給定的四個數，首先從中選兩個數，然後對這兩個數選擇加減乘除這四種操作的一個，得出結果後從剩下的3個數中再選兩個，執行一種操作。如此這樣重復遍歷所有可能即可。

class Solution {

    boolean res = false;
    final double eps = 0.001;

    public boolean judgePoint24(int[] nums) {
        List<Double> arr = new ArrayList<>();
        
 
for(int n: nums) arr.add((double) n);
        helper(arr);
        return res;
    }

    private void helper(List<Double> arr){
        if(res) return;
        if(arr.size() == 1){
            if(Math.abs(arr.get(0) - 24.0) < eps)　// java中double類型判斷是否相等的方法，不能直接用是否等於0判斷
                res = true;
            return;
        }
        for (int i = 0; i < arr.size(); i++) {
            for (int j = 0; j < i; j++) {
                List<Double> next = new ArrayList<>();
                Double p1 = arr.get(i), p2 = arr.get(j);
                next.addAll(Arrays.asList(p1+p2, p1-p2, p2-p1, p1*p2));
                if(Math.abs(p2) > eps)  next.add(p1/p2);
                if(Math.abs(p1) > eps)  next.add(p2/p1);
                
                arr.remove(i);
                arr.remove(j);
                for (Double n: next){
                    arr.add(n);
                    helper(arr);
                    arr.remove(arr.size()-1);
                }
                arr.add(j, p2);
                arr.add(i, p1);
            }
        }
    }
}

註意上面List集合的兩個方法：

void add(int index,
         E element)

E remove(int index)

Parameters:

index - the index of the element to be removed

Returns:

the element previously at the specified position

LeetCode題679 —— 24 Game